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I'm having trouble solving this limit:

$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$

How do I (algebraically) determine if it is positive or negative?

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1. There's nothing to rationalize. 2. If you plot your function, you find that the whole thing goes to $\infty$ at $x=-2$ (an essential singularity). –  J. M. May 16 '11 at 1:26
    
yes, but the question is how do I solve it without plotting? how do I know that it goes to infinity and if it is positive or negative? –  Caleb Jares May 16 '11 at 1:29
    
You know $\lim\limits_{x\to 0^-}\frac1{x^2}$, right? Try to turn your limit into that form. –  J. M. May 16 '11 at 1:30
    
I did not know you could change limits like that. If I change $$\lim_{x \to -2^-}$$ to $$\lim_{x \to 0^-}$$ what must happen to the rest of the function? Is there a rule for this? –  Caleb Jares May 16 '11 at 1:33
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@carble729: as you approach -2 from the left, your denominator $(x+2)$ becomes of the same form as if you were approaching $0$ from the left with the denominator being $x$ (in both cases, the denominator goes to 0, and that's the important part). Hopefully it's more obvious now. –  InterestedGuest May 16 '11 at 1:36
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4 Answers

up vote 8 down vote accepted

Maybe this way of thinking about it will seem a little more intuitive to you:

Let $\varepsilon > 0$, and consider the limit

$$\lim_{x \rightarrow -2^{-}} \frac{1}{(x+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{((-2-\varepsilon)+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon^2}.$$

Now the expression $\frac{1}{\varepsilon^2}$ can be made arbitrarily large by choosing $\varepsilon$ small enough, and so the limit does not exist.

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Thank you, that does make sense (of course DNE being the same thing as + or - infinity - + in this case). When I asked this question, I didn't know that it was positive both ways. Can you explain how to get + or - infinity from the following problem? $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$ –  Caleb Jares May 16 '11 at 2:11
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First, $$\lim_{x \rightarrow 3^+} \frac{x-4}{x-3} = \lim_{x \rightarrow 3^+} \frac{(x-3)-1}{x-3} = \lim_{x \rightarrow 3^+} 1 - \frac{1}{x-3}.$$ Then letting $\varepsilon > 0$, we have $$\lim_{x \rightarrow 3^+} 1 - \frac{1}{x-3} = \lim_{\varepsilon \rightarrow 0} 1 - \frac{1}{(3+\varepsilon) - 3} = \lim_{\varepsilon \rightarrow 0} 1 - \frac{1}{\varepsilon}.$$ Now since $\varepsilon > 0$, $\frac{1}{\varepsilon} > 0$, and this goes to $-\infty$. –  Dane May 16 '11 at 2:23
    
This makes perfect sense! Thank you! –  Caleb Jares May 16 '11 at 2:26
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You asked for an algebraic solution and that's well-covered in other answers, but I'd like to offer a conceptual way of reasoning through it. Let's start with your revised example: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$ As $x$ gets close to $3$, the numerator is getting close to $-1$ and the denominator is getting close to $0$. Something nonzero (rather, not tiny-approaching-zero) divided by something tiny-approaching-zero is going to be big (far from zero), suggesting a limit of $\infty$ or $-\infty$ (that is, suggesting that the limit does not exist, but may be of the specific $\infty$ or $-\infty$ kind of does-not-exist).

Now, how do we tell whether it's $\infty$, $-\infty$, or neither? Well, the numerator of the fraction is getting close to $-1$, so it's negative. The denominator of the fraction is getting close to zero, but specifically as $x\to 3^+$, $x>3$, so $x-3>0$ and the denominator is positive. The fraction is the quotient of a negative number and a positive number, so it's negative and $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}\to-\infty.$$

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As $x \to -2^{-}$, your fraction becomes of form $\frac{1}{\epsilon}$, where $\epsilon$ is an arbitrarily small number, and 1 divided by a 'small' number is clearly a 'large' number (the smaller the denominator, the larger the value of the fraction). I hope you can see it from here.

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I know that I could do it like that, but that's still plugging in values (albeit inside your head). I'm interested to find a way to solve it without plugging in numbers (even if it's in your head) or graphing it. It makes more sense to me if I can understand how the math works in an absolute sense. –  Caleb Jares May 16 '11 at 1:36
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@carble729: It's not really plugging in numbers -- I am assuming that you 'know' that limit of 1 over $x$ where $x$ goes to $0$ is equal to infinity. –  InterestedGuest May 16 '11 at 1:37
    
Yes, but what if the problem is $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$. Approached from the right, it is $-\infty$ and from the left, it is $+\infty$. How do I tell if it is positive or negative infinity without graphing it or plugging in numbers? –  Caleb Jares May 16 '11 at 2:06
    
@cable: make a proper change of variables such that you're only considering limits of the form $u\to0^+$ or $u\to0^-$, as needed for your problem. –  J. M. May 16 '11 at 2:20
    
@cable729: you consider the values: for every $x$ greater than $3$, $x-3$ is positive. For all $x$ between $3$ and $4$, $x-4$ is negative. So for $x$ sufficiently close to $3$, on the right, the fraction is negative (a negative number divided by a positive number). –  Arturo Magidin May 16 '11 at 2:22
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Given any $M > 0$, we have that $\forall x \in \left( -2 - \frac1{\sqrt{M}},-2 \right)$, we have $\frac1{(x+2)^2} > M$.

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Thanks for the answer, but I'm having trouble understanding this (I'm reviewing for a Calc 1 final). I'm not sure what $\forall$ and $\in$ are. Also, I'm not sure what M stands for. –  Caleb Jares May 16 '11 at 1:39
    
what it means is give me any positive number $M$, then if $x$ lies in the interval $\left( -2-\frac1{\sqrt{M}},-2 \right)$, then $\frac1{(x+2)^2} > M$. This implies that the fraction can be made arbitrarily large by getting "sufficiently close" to -$2$. $\forall$ stands for "forall" and $\in$ stands for "belongs to" –  user17762 May 16 '11 at 1:41
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